- #1
Ben Boldt
- 1
- 1
- TL;DR Summary
- Resonance of a cylinder/tube with a hole in the center
The Science Museum of Minnesota has this really cool xylophone hanging from the ceiling, named "Seismofon":
There is a computer connected that receives almost-real-time seismic activity, and it uses that information to play various chords and riffs, etc. So it is always making little random noises basically.
I wanted to reproduce 1 of the tubes, just as a personal project for my own curiosity. You will notice, it is not exactly like a normal xylophone. The resonators are horizontal, with a hole in the CENTER, and the xylophone bar is parallel to the resonator. There is a solenoid that strikes the bar. I have determined that the tubes seen in the photo are 3 inch diameter, and they are very thin-walled. It would be pretty much exactly an exhaust pipe from a pickup truck.
I found an old 3" PVC pipe to experiment with. The inside diameter is 3", and the outside diameter is 3.5". I wanted this pipe to resonate at 440 Hz (Musical Note A4).
I found this website that describes the equations to determine the correct length of resonator versus frequency:
http://supermediocre.org/index.php/2016/04/13/tuning-the-tubes/
Tube closed on one end:
f1 = c / [ 4 (L + 0.61 r) ]
Tube open on both ends or closed on both ends:
f1 = c / [ 2 (L + 0.61 r) ]
f1: fundamental frequency
c: speed of sound
L: length of resonator pipe
r: radius of resonator pipe
At my elevation, the speed of sound is about c = 344.44 m/s.
I want f1 = 435 Hz (I can sand it shorter to tune it up to 440 Hz).
r = 1.5 inches = 0.0381 meters
My assumption, which seems to be wrong, was that the hole in the middle makes it like 2 separate open-ended resonators. So I solved the length using the second equation, then I cut the pipe twice as long as that and put a 3" hole in the center.
[435 Hz] = [344.44 m/s] / [ 2 (L + (0.61 * 0.0381) ) ]
<<solved for L>>
L = 0.3727 meters = 14.67 inches
2L = 29.34 inches
This is what my pipe looks like:
That same website from before, the guy described how he used a speaker and microphone to play a sine wave into his resonator. It starts at a low frequency, and moves higher. The resonator will amplify the sound at its fundamental frequency (f1). I tried this using this website to generate the sweep:
https://onlinesound.net/sweep-tone-generator
I found that my pipe's fundamental frequency was much lower than expected. It is about 377 Hz even though I was shooting for 435 Hz. So I do not think my assumption is correct that I can treat this like 2 open-ended resonators. I did not find any harmonic resonance lower than 377 Hz, so I am feeling like that is the fundamental frequency of this particular pipe.
Looking for clues and ideas, I noticed that if I divide 377Hz / 435Hz, I get a value very close to (√3)/2. The unit circle has the point (1/2, √3/2), seeing as how the hole is half-way down the pipe, maybe that gives us √3/2? It seems like a stretch but I am not sure what to think.
Does anyone have any insight how to how to correctly calculate this pipe length with the hole in the middle?
There is a computer connected that receives almost-real-time seismic activity, and it uses that information to play various chords and riffs, etc. So it is always making little random noises basically.
I wanted to reproduce 1 of the tubes, just as a personal project for my own curiosity. You will notice, it is not exactly like a normal xylophone. The resonators are horizontal, with a hole in the CENTER, and the xylophone bar is parallel to the resonator. There is a solenoid that strikes the bar. I have determined that the tubes seen in the photo are 3 inch diameter, and they are very thin-walled. It would be pretty much exactly an exhaust pipe from a pickup truck.
I found an old 3" PVC pipe to experiment with. The inside diameter is 3", and the outside diameter is 3.5". I wanted this pipe to resonate at 440 Hz (Musical Note A4).
I found this website that describes the equations to determine the correct length of resonator versus frequency:
http://supermediocre.org/index.php/2016/04/13/tuning-the-tubes/
Tube closed on one end:
f1 = c / [ 4 (L + 0.61 r) ]
Tube open on both ends or closed on both ends:
f1 = c / [ 2 (L + 0.61 r) ]
f1: fundamental frequency
c: speed of sound
L: length of resonator pipe
r: radius of resonator pipe
At my elevation, the speed of sound is about c = 344.44 m/s.
I want f1 = 435 Hz (I can sand it shorter to tune it up to 440 Hz).
r = 1.5 inches = 0.0381 meters
My assumption, which seems to be wrong, was that the hole in the middle makes it like 2 separate open-ended resonators. So I solved the length using the second equation, then I cut the pipe twice as long as that and put a 3" hole in the center.
[435 Hz] = [344.44 m/s] / [ 2 (L + (0.61 * 0.0381) ) ]
<<solved for L>>
L = 0.3727 meters = 14.67 inches
2L = 29.34 inches
This is what my pipe looks like:
That same website from before, the guy described how he used a speaker and microphone to play a sine wave into his resonator. It starts at a low frequency, and moves higher. The resonator will amplify the sound at its fundamental frequency (f1). I tried this using this website to generate the sweep:
https://onlinesound.net/sweep-tone-generator
I found that my pipe's fundamental frequency was much lower than expected. It is about 377 Hz even though I was shooting for 435 Hz. So I do not think my assumption is correct that I can treat this like 2 open-ended resonators. I did not find any harmonic resonance lower than 377 Hz, so I am feeling like that is the fundamental frequency of this particular pipe.
Looking for clues and ideas, I noticed that if I divide 377Hz / 435Hz, I get a value very close to (√3)/2. The unit circle has the point (1/2, √3/2), seeing as how the hole is half-way down the pipe, maybe that gives us √3/2? It seems like a stretch but I am not sure what to think.
Does anyone have any insight how to how to correctly calculate this pipe length with the hole in the middle?
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